blocks|key|1933889|text|关于|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1933890|++++function+unique(list)+{
++++++for+(i+=+0;+i<list.length;+i%2B%2B)+{
++++++++for+(j=i%2B1;+j<list.length;+j%2B%2B)+{
++++++++++if+(list[i]+==+list[j])+{
++++++++++++list.splice(j,+1);
++++++++++}
++++++++}
++++++}
++++}|code-block|syntax|javascript|1933891|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>what about</p>

<pre><code>    function unique(list) {
      for (i = 0; i&lt;list.length; i++) {
        for (j=i+1; j&lt;list.length; j++) {
          if (list[i] == list[j]) {
            list.splice(j, 1);
          }
        }
      }
    }
</code></pre>


blocks|key|1934290|text|适用于小列表的另一种方法是模仿sort+%7C+uniq的Unix命令行方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1934291|++++function+unique(a)+{
++++++++return+a.sort().filter(function(value,+index,+array)+{
++++++++++++return+(index+===+0)+%7C%7C+(value+!==+array[index-1]);
++++++++});
++++}|code-block|syntax|javascript|1934292|此函数对参数进行排序，然后筛选结果以忽略与其前身相等的所有项。|1934293|基于键的方法很好，对于大量的项(+O(n+)用于将n项插入到哈希表中，而O(N+log+n)用于对数组进行排序)将具有更好的性能特征。然而，这在小列表中不太可能被注意到。此外，在这个版本中，如果需要，您可以修改它以使用不同的排序或相等函数；使用散列键，您将坚持使用键相等的JavaScripts概念。|1934294|entityMap^0|F|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

<p>An alternative that's suitable for small lists would be to ape the Unix command line approach of <code>sort | uniq</code>:</p>

<pre><code>    function unique(a) {
        return a.sort().filter(function(value, index, array) {
            return (index === 0) || (value !== array[index-1]);
        });
    }
</code></pre>

<p>This function sorts the argument, and then filters the result to omit any items that are equal to their predecessor.</p>

<p>The keys-based approach is fine, and will have better performance characteristics for large numbers of items (O(n) for inserting n items into a hashtable, compared to O(n log n) for sorting the array). However, this is unlikely to be noticeable on small lists. Moreover, with this version you could modify it to use a different sorting or equality function if necessary; with hash keys you're stuck with JavaScripts notion of key equality.</p>


blocks|key|1934478|text|这是一个古老的问题，我知道。然而，它位于一些谷歌搜索的顶部，所以我想补充一句，你可以使用以下命令将@RobHague和@EugeneNaydenov的答案结合起来：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1934479|function+unique(arr)+{
++const+u+=+{};
++return+arr.filter((v)+=>+{
++++return+u[v]+=+!u.hasOwnProperty(v);
++});
};|code-block|syntax|javascript|1934480|1934481|function+unique(arr)+{
++const+u+=+{};
++return+arr.filter((v)+=>+{
++++return+u[v]+=+(v+!==+undefined+&&+!u.hasOwnProperty(v));
++});
};|1934482|你可以在这里试用这个解决方案：https://jsfiddle.net/s8d14v5n/|offset|length|1934483|entityMap|0|LINK|mutability|MUTABLE|url|https://jsfiddle.net/s8d14v5n/^0|0|0|0|0|F|U|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|H|3|I|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|J|3|K|5|6|7|Z|8|@]|9|@$L|10|M|11|1|12]]|A|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|U]]]]

<p>This is an old question, I know. However, it is at the top of some google searches, so I wanted to add that you can combine the answers from @RobHague and @EugeneNaydenov using the following:</p>

<pre><code>function unique(arr) {
  const u = {};
  return arr.filter((v) =&gt; {
    return u[v] = !u.hasOwnProperty(v);
  });
};
</code></pre>

<p>You can also ignore <em>undefined</em> values (often handy) by adding:</p>

<pre><code>function unique(arr) {
  const u = {};
  return arr.filter((v) =&gt; {
    return u[v] = (v !== undefined &amp;&amp; !u.hasOwnProperty(v));
  });
};
</code></pre>

<p>You can play with this solution here: <a href="https://jsfiddle.net/s8d14v5n/" rel="nofollow noreferrer">https://jsfiddle.net/s8d14v5n/</a></p>


blocks|key|1720909|text|如果你输入整数参数(uniq(1,2,3)+=>+'1'，'2'，'3‘)，使用Object.keys就会得到字符串。下面是一个使用Array.reduce的示例：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1720910|function+uniq(list)+{
++++return+list.reduce((acc,+d)+=>+acc.includes(d)+?+acc+:+acc.concat(d),+[]);
}|code-block|syntax|javascript|1720911|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<p>Using Object.keys will give you strings if you put in integer arguments (uniq([1,2,3]) => ['1','2','3']. Here's one with Array.reduce:</p>

<pre><code>function uniq(list) {
    return list.reduce((acc, d) =&gt; acc.includes(d) ? acc : acc.concat(d), []);
}
</code></pre>


blocks|key|1720967|text|这应该适用于任何东西，而不仅仅是字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1720968|export+const+getUniqueList+=++(a:+Array<any>)+:+Array<any>+=>+{

++const+set+=+new+Set<any>();

++for(let+v+of+a){
++++++set.add(v);
++}

++return+Array.from(set);

};|code-block|syntax|javascript|1720969|以上内容可以简化为：|1720970|export+const+getUniqueValues+=+(a:+Array<any>)+=>+{
+++return+Array.from(new+Set(a));
};|1720971|:)|1720972|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<p>This should work with anything, not just strings:</p>

<pre><code>export const getUniqueList =  (a: Array&lt;any&gt;) : Array&lt;any&gt; =&gt; {

  const set = new Set&lt;any&gt;();

  for(let v of a){
      set.add(v);
  }

  return Array.from(set);

};
</code></pre>

<p>the above can just be reduced to:</p>

<pre><code>export const getUniqueValues = (a: Array&lt;any&gt;) =&gt; {
   return Array.from(new Set(a));
};
</code></pre>

<p>:)</p>


blocks|key|1721028|text|最好的方法似乎是使用ES6和Set。根据fiddle的说法，单行速度*比上面更快|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1721029|​|1721030|const+myList+=+[1,4,5,1,2,4,5,6,7];
const+unique+=+[...new+Set(myList)];
++++
console.log(unique);|code-block|syntax|javascript|1721031|1721032|*在safari中测试|1721033|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/8gzmtao5/^0|K|6|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|10|8|@]|9|@]|C|$I|J]]|$1|K|3|E|5|6|7|11|8|@]|9|@]|C|$]]|$1|L|3|M|5|6|7|12|8|@]|9|@]|C|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|O|$P|$5|Q|R|S|C|$T|U]]]]

<p>The best method seem to be using ES6 and Set. Single line and faster* than above according to <a href="http://jsfiddle.net/8gzmtao5/" rel="noreferrer">fiddle</a></p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>    
const myList = [1,4,5,1,2,4,5,6,7];
const unique = [...new Set(myList)];
    
console.log(unique);</code></pre>
</div>
</div>
</p>

<p>*tested in safari</p>


blocks|key|1721074|text|我发现其他的答案是相当复杂的，在我看来是徒劳的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1721075|在推送之前，我们可以使用数组的indexOf方法来验证其中是否存在项目：|offset|length|1721076|​|1721077|const+duplicated_values+=+['one',+'one',+'one',+'one',+'two',+'three',+'three',+'four'];
const+unique_list+=+[];

duplicated_values.forEach(value+=>+{
++if+(unique_list.indexOf(value)+===+-1)+{
++++unique_list.push(value);
++}
});

console.log(unique_list);|code-block|syntax|javascript|1721078|1721079|这也适用于任何类型的变量，甚至对象(给定标识符实际上引用了相同的实体，只是等效的对象不被视为相同的)。|1721080|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf^0|0|F|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@$D|Z|E|10|1|11]]|A|$]]|$1|F|3|G|5|6|7|12|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|13|8|@]|9|@]|A|$K|L]]|$1|M|3|G|5|6|7|14|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|15|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

<p>I find the other answers to be rather complicated for no gain that I can see.</p>
<p>We can use the <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/indexOf" rel="nofollow noreferrer">indexOf</a> method of the Array to verify if an item exists in it before pushing:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const duplicated_values = ['one', 'one', 'one', 'one', 'two', 'three', 'three', 'four'];
const unique_list = [];

duplicated_values.forEach(value =&gt; {
  if (unique_list.indexOf(value) === -1) {
    unique_list.push(value);
  }
});

console.log(unique_list);</code></pre>
</div>
</div>
</p>
<p>That will work with any type of variable as well, even objects (given the identifier actually reference the same entity, merely equivalent objects are not seen as the same).</p>


blocks|key|1934700|text|要获得唯一的对象，可以使用JSON.stringify和JSON.parse|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1934701|​|1934702|const+arr+=+[{test:+"a"},+{test:+"a"}];
const+unique+=+Array.from(new+Set(arr.map(JSON.stringify))).map(JSON.parse);
console.log(unique);|code-block|syntax|javascript|1934703|1934704|entityMap^0|D|E|S|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|V|8|@]|D|@]|E|$K|L]]|$1|M|3|G|5|6|7|W|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|O|$]]

<p>To get unique objects, you can use <code>JSON.stringify</code> and <code>JSON.parse</code>:</p>
<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>const arr = [{test: "a"}, {test: "a"}];
const unique = Array.from(new Set(arr.map(JSON.stringify))).map(JSON.parse);
console.log(unique);</code></pre>
</div>
</div>
</p>


<p>In my CouchDB reduce function I need to reduce a list of items to the unique ones.</p>

<p><em>Note: In that case it's ok to have a list, it will be a small number of items of string type.</em></p>

<p>My current way is to set keys of a object, then return the keys of that object
since the place the code can't use things like <code>_.uniq</code> for example.</p>

<p>I'd like to find a more elegant way to spell it than this.</p>

<pre><code>function(keys, values, rereduce) {
  // values is a Array of Arrays
  values = Array.concat.apply(null, values);
  var uniq = {};
  values.forEach(function(item) { uniq[item] = true; });
  return Object.keys(uniq);
}
</code></pre>


How to create a list of unique items in JavaScript?

JSON

Unix

在我的CouchDB reduce函数中，我需要将一个条目列表缩减为唯一的条目。注意:在这种情况下，可以有一个列表，它将是一小部分字符串类型的项目。我目前的方法是设置一个对象的键，然后返回该对象的键，因为代码不能使用像_.uniq这样的东西。我想找到一种比这个更优雅的拼写方式。function(keys, values...

问如何在JavaScript中创建唯一项的列表？
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